# Doesn't light have to be a wave since it has a wave length? (In contrast to the wave–particle duality) [duplicate]

We learned in school that photons behave partly like a wave and partly like a particle, so they can be both at the same time. We also learned that light consists of alternating electric and magnetic fields that travel through empty space as seen in the picture below. So if you can always name the specific wave length of a quantum object, doesn't that necessarily imply that this object must be a wave?

• Possible duplicate of Why do we think of light as a wave? – QuantumBrick Aug 31 '16 at 19:25
• A 500 nm photon has a frequency of more than 600 trillion oscillations per second! As it travels along at the speed of light it completes one of these oscillations in a very short distance of 500 nanometers. We call it a wavelength for convenience but it doesn't necessarily have anything to do with waves. At least it has not been proven to be a wave anymore than an oscillating photon particle. – Bill Alsept Aug 31 '16 at 20:54

There is no conflict with the wave-particle duality. That duality just states that light has particle properties as well as wave properties. It has a wave-length, forms usual difraction and refraction patterns, and in most relevant macroscopic cases, it is well described by a wave equation. So, sure, it's a wave. That said, it also has particle-like properties.

Sometimes you read stuff like "the object sometimes behaves like a particle and sometimes like a wave", but that is just nonsense. The fact is that a classical particle and a classical wave are just not the correct description for a quantum object. In that sense, the photon is neither a wave nor a particle. It's something different (actually, the best description we have says that it is just an excitation of a quantum field).

The story that light is an alternating electric and magnetic field is the picture that you get when you interpret the classical equations of electromagnetism. There is nothing wrong with that, but you have to keep in mind that classical electromagnetism is an effective theory, which means that it is only a good approximation to nature on a certain scale. It just isn't the full picture. In a relativistic picture, there is no real distinction between magnetic and electric fields and you only have one field, the electromagnetic one. That field also doesn't look like your picture (I can't tell you how it looks like, I can only draw other pictures).

You can draw a lot of pretty pictures and classical analogies, but none will capture all aspects of the photon.

No, unfortunately not. We could still say sometimes it acts like a wave, and then we treat it as a wave, with all the wave terms, not just wavelength but frequency and amplitude as well.

The problem is that particles also have a wavelength, (a extremely short one) but I don't know if you have covered that yet in your course.

At the end of the 19th century, light was thought to consist of waves of electromagnetic fields which propagated according to Maxwell’s equations, while matter was thought to consist of localized particles. In 1900, this division was exposed to doubt, when, investigating the theory of black body thermal radiation, Max Planck proposed that light is emitted in discrete quanta of energy. It was thoroughly challenged in 1905.

Extending Planck's investigation in several ways, including its connection with the photoelectric effect, Albert Einstein proposed that light is also propagated and absorbed in quanta. Light quanta are now called photons. These quanta would have an energy given by the Planck–Einstein relation:

$$E=\frac{h}{\nu }$$

and a momentum

$$p={\frac {E}{c}} = {\frac {h}{\lambda }}$$

where $$ν$$ (lowercase Greek letter nu) and $$λ$$ (lowercase Greek letter lambda) denote the frequency and wavelength of the light, c the speed of light, and $$h$$ Planck’s constant.

One particular experiment that can only be explained by treating a photon as a "particle" is the Photoelectric Effect

Image Source: By Wolfmankurd - en:Inkscape, CC BY-SA 3.0,

The red wavy lines coming in from the left represent photons (as particles), which have too low a frequency (not enough energy) to free electrons from a metal surface. The green wavy lines are of a higher frequency and do have enough energy to knock electrons of the metal surface.

If light was a wave, this process would take a very long time to release any photons, but we know it happens as soon as you shine a light of the correct frequency ( or higher) on the plate.

The photoelectric effect is a phenomenon in physics. The effect is based on the idea that electromagnetic radiation is made of a series of particles called photons. When a photon hits an electron on a metal surface, the electron can be emitted. The emitted electrons are called photoelectrons.

The photoelectric effect has helped physicists understand the quantum nature of light and electrons. The concept of wave–particle duality was developed because of the photoelectric effect.

One way to think about wave-particle duality is to say that it's a duality between wavevector-$\mathbf{k}$ wave and momentum-$\hbar\mathbf{k}$ particle interpretations. (The Dirac constant $\hbar$ is $h/2\pi$, with $h$ the Planck constant.) The wavevector of a wave is its wavenumber $k=2\pi/\lambda$ (with $\lambda$ the wavelength) multiplied by a unit vector in its direction of propagation, as in the Ansatz $\exp i\mathbf{k}\cdot\mathbf{x}=\exp ikx\cos\theta$. The momentum of this Ansatz can be computed as its eigenvalue under the momentum operator $-i\hbar\boldsymbol{\nabla}$, which is $\hbar k$. Taking moduli, $p=\hbar k=2\pi\hbar/\lambda=h/\lambda$, the de Broglie hypothesis. For photons, we also have a result due to Planck, $E=hf=\hbar\omega$ with $f=c/\lambda$ the linear frequency and $\omega=2\pi f$ the angular or circular frequency.